Prove monotonic function on $mathbb{R}$ under given condition
$begingroup$
Let $f(x)=e^{x}-x^2-ax$
(a) Prove when $aleq 2-2ln(2)$ , $f(x)$ is monotonic function on $mathbb{R},(a,+infty)$
(b) Given when $x>0$, $f(x)geq 1-x$ always true. Find the range of $a$.
This is a question from middle school exam paper on the topic related to monotonic function. Can anyone solve this? I was wondering why middle school exam contains such a weird question.
derivatives monotone-functions
edited Jan 18 at 15:37
idriskameni
749321
asked Jan 18 at 14:07
CColaCCola
1317
$endgroup$
add a comment |
$begingroup$
Let $f(x)=e^{x}-x^2-ax$
(a) Prove when $aleq 2-2ln(2)$ , $f(x)$ is monotonic function on $mathbb{R},(a,+infty)$
(b) Given when $x>0$, $f(x)geq 1-x$ always true. Find the range of $a$.
This is a question from middle school exam paper on the topic related to monotonic function. Can anyone solve this? I was wondering why middle school exam contains such a weird question.
derivatives monotone-functions
edited Jan 18 at 15:37
idriskameni
749321
asked Jan 18 at 14:07
CColaCCola
1317
$endgroup$
add a comment |
$begingroup$
Let $f(x)=e^{x}-x^2-ax$
(a) Prove when $aleq 2-2ln(2)$ , $f(x)$ is monotonic function on $mathbb{R},(a,+infty)$
(b) Given when $x>0$, $f(x)geq 1-x$ always true. Find the range of $a$.
This is a question from middle school exam paper on the topic related to monotonic function. Can anyone solve this? I was wondering why middle school exam contains such a weird question.
derivatives monotone-functions
edited Jan 18 at 15:37
idriskameni
749321
asked Jan 18 at 14:07
CColaCCola
1317
$endgroup$
Let $f(x)=e^{x}-x^2-ax$
(a) Prove when $aleq 2-2ln(2)$ , $f(x)$ is monotonic function on $mathbb{R},(a,+infty)$
(b) Given when $x>0$, $f(x)geq 1-x$ always true. Find the range of $a$.
This is a question from middle school exam paper on the topic related to monotonic function. Can anyone solve this? I was wondering why middle school exam contains such a weird question.
derivatives monotone-functions
derivatives monotone-functions
edited Jan 18 at 15:37
idriskameni
749321
asked Jan 18 at 14:07
CColaCCola
1317
edited Jan 18 at 15:37
idriskameni
749321
asked Jan 18 at 14:07
CColaCCola
1317
edited Jan 18 at 15:37
idriskameni
749321
edited Jan 18 at 15:37
idriskameni
749321
edited Jan 18 at 15:37
idriskameni
749321
749321
asked Jan 18 at 14:07
CColaCCola
1317
asked Jan 18 at 14:07
CColaCCola
1317
asked Jan 18 at 14:07
CColaCCola
1317
1317
add a comment |
add a comment |
1 Answer
1
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$begingroup$
Hint:
Try to use $x=ln(y)$.
Then you have $f(y) = y-ln(y)^2-aln(y)$.
Then $f'(y)= 1-frac{2}{y}ln(y)-frac{a}{y}$.
answered Jan 18 at 14:44
idriskameniidriskameni
749321
$endgroup$
add a comment |
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1 Answer
1
active
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1 Answer
1
active
oldest
votes
active
oldest
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active
oldest
votes
$begingroup$
Hint:
Try to use $x=ln(y)$.
Then you have $f(y) = y-ln(y)^2-aln(y)$.
Then $f'(y)= 1-frac{2}{y}ln(y)-frac{a}{y}$.
answered Jan 18 at 14:44
idriskameniidriskameni
749321
$endgroup$
add a comment |
$begingroup$
Hint:
Try to use $x=ln(y)$.
Then you have $f(y) = y-ln(y)^2-aln(y)$.
Then $f'(y)= 1-frac{2}{y}ln(y)-frac{a}{y}$.
answered Jan 18 at 14:44
idriskameniidriskameni
749321
$endgroup$
add a comment |
$begingroup$
Hint:
Try to use $x=ln(y)$.
Then you have $f(y) = y-ln(y)^2-aln(y)$.
Then $f'(y)= 1-frac{2}{y}ln(y)-frac{a}{y}$.
answered Jan 18 at 14:44
idriskameniidriskameni
749321
$endgroup$
Hint:
Try to use $x=ln(y)$.
Then you have $f(y) = y-ln(y)^2-aln(y)$.
Then $f'(y)= 1-frac{2}{y}ln(y)-frac{a}{y}$.
answered Jan 18 at 14:44
idriskameniidriskameni
749321
answered Jan 18 at 14:44
idriskameniidriskameni
749321
answered Jan 18 at 14:44
idriskameniidriskameni
749321
answered Jan 18 at 14:44
idriskameniidriskameni
749321
749321
add a comment |
add a comment |
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